Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of $\theta $, where $0

Question

Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of

\theta

, where

0 < \theta < \frac { \pi } { 2 }

.

\sqrt { 36 - 9 x ^ { 2 } } , x = 2 \cos \theta

A)2 cos $\theta $
B)36 sin $\theta $
C)- 2 sin$\theta $
D)36 cos $\theta $
E)6 sin $\theta $

Quizplus · Accepted Answer

Given $x = 2\cos\theta$, we substitute $x$ in the expression $\sqrt{36 - 9x^2}$. This becomes $\sqrt{36 - 9(2\cos\theta)^2}$ = $\sqrt{36 - 36\cos^2\theta}$ = $\sqrt{36(1 - \cos^2\theta)}$. Since $\sin^2\theta + \cos^2\theta = 1$, we have $\sqrt{36\sin^2\theta}$ = $6\sin\theta$, where $0 < \theta < \frac{\pi}{2}$.

Use the Trigonometric Substitution to Rewrite the Algebraic Expression as a Trigonometric